a game involves a six - sided die. if the player rolls a 1 or 2, they win $10. if they roll a 3 or 4, they…

a game involves a six - sided die. if the player rolls a 1 or 2, they win $10. if they roll a 3 or 4, they win $5. if they roll a 5 or 6, they lose $5. what is the expected value of the game?\na $1.67\nb $2.50\nc $3.33\nd $5.00
Answer
Explanation:
Step1: Calculate probabilities
The probability of rolling a 1 or 2, a 3 or 4, or a 5 or 6 is $\frac{2}{6}=\frac{1}{3}$ for each case.
Step2: Calculate products of outcomes and probabilities
For winning $10 (probability $\frac{1}{3}$), the product is $10\times\frac{1}{3}=\frac{10}{3}$. For winning $5 (probability $\frac{1}{3}$), the product is $5\times\frac{1}{3}=\frac{5}{3}$. For losing $5 (probability $\frac{1}{3}$), the product is $(- 5)\times\frac{1}{3}=-\frac{5}{3}$.
Step3: Calculate expected value
The expected - value $E(X)$ is the sum of the products: $E(X)=\frac{10}{3}+\frac{5}{3}-\frac{5}{3}=\frac{10}{3}\approx3.33$.
Answer:
C. $3.33